Question
Show that $\sin100^\circ-\sin10^\circ$ is positive.
✓
Answer
We have, $\sin100^\circ-\sin10^\circ$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\times100^\circ-\frac{1}{\sqrt{2}}\times\cos100^\circ\Big)$ $\big[$Multiplying and dividing by $\sqrt{1^2+1^2}$ i.e., by $\sqrt{2}\big]$
$=\sqrt{2}(\cos45^\circ\times\sin100^\circ-\sin45^\circ\times\cos100^\circ)$
$=\sqrt{2}(\sin100^\circ\times\cos45^\circ-\cos100^\circ\times\sin45^\circ)$
$=\sqrt{2}(\sin(100^\circ-45^\circ))$
$=\sqrt{2}\sin55^\circ,$ which is positive real number. $[\because\sin\theta$ is positive in first quadrant$]$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\times100^\circ-\frac{1}{\sqrt{2}}\times\cos100^\circ\Big)$ $\big[$Multiplying and dividing by $\sqrt{1^2+1^2}$ i.e., by $\sqrt{2}\big]$
$=\sqrt{2}(\cos45^\circ\times\sin100^\circ-\sin45^\circ\times\cos100^\circ)$
$=\sqrt{2}(\sin100^\circ\times\cos45^\circ-\cos100^\circ\times\sin45^\circ)$
$=\sqrt{2}(\sin(100^\circ-45^\circ))$
$=\sqrt{2}\sin55^\circ,$ which is positive real number. $[\because\sin\theta$ is positive in first quadrant$]$
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